2026 HSC CHAT (9 Viewers)

NotBamboo

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ok so here's my working
so if you draw up x^2 +2x+3 you can see domain is all real x because it never touches 0 and below due to the square root sign so that's domain

onto the range, so just for the quadratic we can see the minimum value/turning point is -1 (just remembered formula is -b/2a whoops). We use the min value bc the smaller the x value the larger the y value for the function.
Using -1 we sub it into the 1/sqrt of x^2 +2x+3 and our maximum value is 0.7071...
as the quadratic HAS to be greater than 0 (bc you can't do 1/0) and also because when doing square root functions you don't get negative (only outputs positive apparently)
range is : 0 < y </ (less than equal to) 0.7071...

(does this make sense or do I need a diagram?)
GOOD STUFF KRYSTAL IM PROUD OF U <3
yeah heres my full 12 pages of working out that u did in like 5 sentences
Domain & Range help for my bestie | Bored Of Studies
 

NotBamboo

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I just find the range of the quadratic, then sqrt and reciprocate it
like the quadratic is y ≥ 2
sqrt it so y ≥ sqrt2
and then when u take the recpirocal u get 0 ≤ y ≤ 1/sqrt2
huh???
can u explain it for me plssss
how did u get 2?
 

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